我如何连接两个std::向量?
当前回答
在c++ 11中,我更喜欢将向量b附加到a:
std::move(b.begin(), b.end(), std::back_inserter(a));
当a和b不重叠时,b不会再被用到。
这是std::move from <algorithm>,而不是通常的std::move from <utility>。
其他回答
你应该使用vector::insert
v1.insert(v1.end(), v2.begin(), v2.end());
我已经实现了这个函数,它连接任何数量的容器,从右值引用移动和复制
namespace internal {
// Implementation detail of Concatenate, appends to a pre-reserved vector, copying or moving if
// appropriate
template<typename Target, typename Head, typename... Tail>
void AppendNoReserve(Target* target, Head&& head, Tail&&... tail) {
// Currently, require each homogenous inputs. If there is demand, we could probably implement a
// version that outputs a vector whose value_type is the common_type of all the containers
// passed to it, and call it ConvertingConcatenate.
static_assert(
std::is_same_v<
typename std::decay_t<Target>::value_type,
typename std::decay_t<Head>::value_type>,
"Concatenate requires each container passed to it to have the same value_type");
if constexpr (std::is_lvalue_reference_v<Head>) {
std::copy(head.begin(), head.end(), std::back_inserter(*target));
} else {
std::move(head.begin(), head.end(), std::back_inserter(*target));
}
if constexpr (sizeof...(Tail) > 0) {
AppendNoReserve(target, std::forward<Tail>(tail)...);
}
}
template<typename Head, typename... Tail>
size_t TotalSize(const Head& head, const Tail&... tail) {
if constexpr (sizeof...(Tail) > 0) {
return head.size() + TotalSize(tail...);
} else {
return head.size();
}
}
} // namespace internal
/// Concatenate the provided containers into a single vector. Moves from rvalue references, copies
/// otherwise.
template<typename Head, typename... Tail>
auto Concatenate(Head&& head, Tail&&... tail) {
size_t totalSize = internal::TotalSize(head, tail...);
std::vector<typename std::decay_t<Head>::value_type> result;
result.reserve(totalSize);
internal::AppendNoReserve(&result, std::forward<Head>(head), std::forward<Tail>(tail)...);
return result;
}
vector<int> v1 = {1, 2, 3, 4, 5};
vector<int> v2 = {11, 12, 13, 14, 15};
copy(v2.begin(), v2.end(), back_inserter(v1));
如果希望能够简洁地连接向量,可以重载+=运算符。
template <typename T>
std::vector<T>& operator +=(std::vector<T>& vector1, const std::vector<T>& vector2) {
vector1.insert(vector1.end(), vector2.begin(), vector2.end());
return vector1;
}
然后你可以这样调用它:
vector1 += vector2;
对于提供push_back (string, vector, deque,…)功能的容器:
Std::copy(Std::begin(input), Std::end(input), Std::back_inserter(output))
and
对于提供insert (map, sets)功能的容器:
Std::copy(Std::begin(input), Std::end(input), Std::inserter(output, output.end()))
推荐文章
- 什么是“参数依赖查找”(又名ADL,或“Koenig查找”)?
- 公共朋友交换成员函数
- 如何在Go中使用c++
- 如何在c#中连接列表?
- 自定义c++分配器的引人注目的例子?
- RAII和c++中的智能指针
- 如何构建和使用谷歌TensorFlow c++ api
- 断言是邪恶的吗?
- 下面这些短语在c++中是什么意思:0 -,default-和value-initialization?
- 在STL地图中,使用map::insert比[]更好吗?
- C++ Linux的想法?
- 如何为Fedora安装g++ ?
- Std::cin输入空格?
- c++标准是否要求iostreams的性能很差,或者我只是在处理一个糟糕的实现?
- gcc在哪里查找C和c++头文件?